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![](https://rs.olm.vn/images/avt/0.png?1311)
C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)
C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)
C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)
C = \(\frac{-233}{135}\)
D = \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)
D = \(-4.\frac{12}{13}\)
D = \(\frac{-48}{13}\)
E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)
E = \(5.4-4.3+5-0,3.20\)
E = \(20-12+5-6\)
E = \(8+\left(-1\right)\)
E = \(7\)
F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\)
F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)
F = \(\frac{-11}{12}\)
Chúc cậu hk tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)
\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)
\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:Ta có:
2800=(28)100=256100(1)
3600=(36)100=729100(2)
Từ (1),(2)
Ta có: 256100<729100(Vì 256<729)
=>2800<3600
Mình làm bài 1 thôi nhé!
Ta có :2^800=2^2.400=(2^2)^400=4^400=4^4.100=(4^4)^100=256^100
3^600=3^2.300=(3^2)^300=9^300=9^3.100=(9^3)^100=729^100
Vì 256^100<729^100 nên 2^800<3^600
^ là mũ nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)
\({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)
Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);
b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)
\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)
Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)
c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)
\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)
Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).
d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)
Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
3. \(\left(\frac{1}{2^5}\right)^{25}=\left(\frac{1^5}{2^5}\right)^{25}=\left[\left(\frac{1}{2}\right)^5\right]^{25}=\left(\frac{1}{2}\right)^{125}\)
\(\left(\frac{1}{3^{25}}\right)^5=\left(\frac{1^{25}}{3^{25}}\right)^5=\left[\left(\frac{1}{3}\right)^{25}\right]^5=\left(\frac{1}{3}\right)^{125}\)
Vì \(\frac{1}{2}>\frac{1}{3}\Rightarrow\left(\frac{1}{2^5}\right)^{25}>\left(\frac{1}{3^{25}}\right)^5\)
1. \(3^{800}=\left(3^8\right)^{100}=6561^{100}\)
\(5^{500}=\left(5^5\right)^{100}=3125^{100}\)
Vì \(6561>3125\Rightarrow3^{800}>5^{500}\)
2. \(\left(-2\right)^{3000}=\left[\left(-2\right)^3\right]^{1000}=\left(-8\right)^{1000}\)
\(\left(-3\right)^{2000}=\left[\left(-3\right)^2\right]^{1000}=9^{1000}\)
Vì \(-8< 9\Rightarrow\left(-2\right)^{3000}< \left(-3\right)^{2000}\)